AP® Chemistry review sheet from Aim for Five (aimforfive.com/chem/units/9/9-9)
Unit 9 · Topic 9.9
9.9 Cell Potential and Free Energy
A cell's standard potential comes from a table of standard reduction potentials: E°cell = E°(cathode) − E°(anode). A positive E°cell means the redox reaction is thermodynamically favored, and ΔG° = −nFE° converts the voltage into a free energy change.
Key terms
- standard reduction potential
- E°cell
- ΔG° = −nFE°
- Faraday constant (F)
- n (moles of electrons)
Standard reduction potentials
Every half-reaction is listed in tables as a reduction, with a standard reduction potential, E°, in volts. The more positive E° is, the more strongly that species pulls electrons, meaning it's easier to reduce. Values are measured relative to hydrogen, 2H⁺ + 2e⁻ → H₂, which is set at 0.00 V.
Strong oxidizing agents (like Cl₂ and Ag⁺) sit near the top with positive E° values. Species that are hard to reduce (like Zn²⁺ and Al³⁺) have negative E° values, which means their metals are easily oxidized.
| Reduction half-reaction | E° (V) |
|---|---|
| Cl₂(g) + 2e⁻ → 2Cl⁻(aq) | +1.36 |
| Ag⁺(aq) + e⁻ → Ag(s) | +0.80 |
| Fe³⁺(aq) + e⁻ → Fe²⁺(aq) | +0.77 |
| Cu²⁺(aq) + 2e⁻ → Cu(s) | +0.34 |
| 2H⁺(aq) + 2e⁻ → H₂(g) | 0.00 |
| Ni²⁺(aq) + 2e⁻ → Ni(s) | −0.25 |
| Zn²⁺(aq) + 2e⁻ → Zn(s) | −0.76 |
| Al³⁺(aq) + 3e⁻ → Al(s) | −1.66 |
Calculating E°cell
Identify which species is reduced (cathode) and which is oxidized (anode). Then E°cell = E°(cathode) − E°(anode), using both values exactly as they appear in the reduction table.
E° is an intensive property: it doesn't depend on how much reacts. When you multiply a half-reaction by 2 to balance electrons, you do not multiply its E°.
For a galvanic cell built from two half-cells, the one with the more positive E° will be the cathode, which always gives a positive E°cell.
Linking E° to ΔG° and K
ΔG° = −nFE°, where n is the moles of electrons transferred in the balanced equation and F is the Faraday constant, 96,485 C per mole of electrons. Since 1 V = 1 J/C, the product nFE° comes out in joules.
A positive E° gives a negative ΔG°, so the reaction is thermodynamically favored and can run as a galvanic cell. A negative E° gives a positive ΔG°: the reaction is unfavored and needs an external voltage (electrolysis).
Combined with ΔG° = −RT ln K, a positive E° also means K > 1.
| E°cell | ΔG° | K | The reaction |
|---|---|---|---|
| Positive | Negative | Greater than 1 | Favored; works as a galvanic cell |
| Zero | Zero | Equal to 1 | At equilibrium under standard conditions |
| Negative | Positive | Less than 1 | Unfavored; needs an external power source |
Predicting whether a metal reacts with an ion
A metal will reduce an ion (and be oxidized) if the ion's reduction potential is higher than that of the metal's own ion. Zinc metal dropped into Cu²⁺ solution gets coated with copper, because +0.34 V is higher than −0.76 V. Copper metal in Zn²⁺ solution does nothing. Copper doesn't dissolve in 1 M HCl either, because H⁺ (0.00 V) has a lower reduction potential than Cu²⁺ (+0.34 V), but zinc does, releasing H₂ gas.
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
E°cell and ΔG° for the zinc–copper cell
For Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), use the table to find E°cell and ΔG°.
Show the solutionHide the solution
- Step 1: Cu²⁺ is reduced, so the cathode is Cu²⁺/Cu (E° = +0.34 V). Zn is oxidized, so the anode is Zn²⁺/Zn (E° = −0.76 V).
- Step 2: E°cell = E°(cathode) − E°(anode) = 0.34 − (−0.76) = +1.10 V.
- Step 3: Two electrons are transferred (Zn → Zn²⁺ + 2e⁻), so n = 2.
- Step 4: ΔG° = −nFE° = −(2)(96,485 C/mol)(1.10 V) = −212,000 J/mol = −212 kJ/mol.
- Step 5: E° is positive and ΔG° is negative, so the reaction is thermodynamically favored.
Answer: E°cell = +1.10 V; ΔG° ≈ −212 kJ/mol
- Example 2Calculator allowed
Trap: don't multiply E° by a coefficient
For Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s), find E°cell, ΔG° and K at 298 K.
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- Step 1: Cathode (reduction): Ag⁺ + e⁻ → Ag, E° = +0.80 V. It's multiplied by 2 to balance electrons, but E° stays +0.80 V. Doubling it to 1.60 V is the trap.
- Step 2: Anode (oxidation): Cu → Cu²⁺ + 2e⁻; its reduction E° is +0.34 V.
- Step 3: E°cell = 0.80 − 0.34 = +0.46 V.
- Step 4: n = 2 (two electrons move from one Cu to two Ag⁺). ΔG° = −(2)(96,485)(0.46) = −88,800 J/mol ≈ −89 kJ/mol.
- Step 5: ln K = −ΔG°/RT = 88,800 ÷ (8.314 × 298) = 35.8, so K ≈ 4 × 10¹⁵. The reaction goes essentially to completion.
Answer: E°cell = +0.46 V; ΔG° ≈ −89 kJ/mol; K ≈ 4 × 10¹⁵
- Example 3Calculator allowed
Is the reverse reaction favored?
Is Cu(s) + Zn²⁺(aq) → Cu²⁺(aq) + Zn(s) thermodynamically favored under standard conditions? What would be needed to make it happen?
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- Step 1: Here Zn²⁺ is reduced (cathode, −0.76 V) and Cu is oxidized (anode, +0.34 V).
- Step 2: E°cell = −0.76 − 0.34 = −1.10 V. Negative.
- Step 3: ΔG° = −(2)(96,485)(−1.10) = +212 kJ/mol. Positive, so not favored.
- Step 4: To make it happen, you'd need an electrolytic cell with an external voltage greater than 1.10 V.
Answer: Not favored (E° = −1.10 V, ΔG° = +212 kJ/mol); it requires an external voltage above 1.10 V.
Common mistakes
- Multiplying E° by the coefficient used to balance electrons.
- Flipping the sign of the anode's E° and then also subtracting it. Use E°(cathode) − E°(anode) with table values, or flip the anode sign and add, but not both.
- Using the wrong n. It's the total electrons transferred in the balanced overall equation.
- Leaving ΔG° in joules when kJ is asked for, or the reverse.
On the exam
- Expect to pick the cathode and anode from a table, calculate E°cell, and state whether the reaction is favored. Showing ΔG° = −nFE° with n clearly identified is a common scoring point.
- You may be asked which of several metals would react with a given ion: the reaction is favored when the ion's reduction potential is higher than that of the metal's ion.
Connected topics
Videos
Check yourself
5 questions on 9.9 Cell Potential and Free Energy. Pick an answer to see if you got it, and why.
Given Ag⁺(aq) + e⁻ → Ag(s), E° = +0.80 V, and Cu²⁺(aq) + 2e⁻ → Cu(s), E° = +0.34 V, what is E°cell for the galvanic cell based on Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s)?
| Half-reaction | E° (V) |
|---|---|
| Au³⁺(aq) + 3e⁻ → Au(s) | +1.50 |
| Ag⁺(aq) + e⁻ → Ag(s) | +0.80 |
| Cu²⁺(aq) + 2e⁻ → Cu(s) | +0.34 |
| Pb²⁺(aq) + 2e⁻ → Pb(s) | −0.13 |
| Fe²⁺(aq) + 2e⁻ → Fe(s) | −0.44 |
| Zn²⁺(aq) + 2e⁻ → Zn(s) | −0.76 |
| Mg²⁺(aq) + 2e⁻ → Mg(s) | −2.37 |
Standard reduction potentials at 25 °C
Which of the following metals would react with 1 M Cu²⁺(aq) but not with 1 M Fe²⁺(aq) under standard conditions?
Which two half-reactions from the table would make a galvanic cell with the largest standard cell potential, and what is that potential?
A galvanic cell is built with a zinc electrode in 1.0 M Zn(NO₃)₂ and a copper electrode in 1.0 M Cu(NO₃)₂, at 25 °C. The electrodes are joined by a wire, and the solutions are joined by a salt bridge containing KNO₃(aq).
Cu²⁺(aq) + 2e⁻ → Cu(s) E° = +0.34 V
Zn²⁺(aq) + 2e⁻ → Zn(s) E° = −0.76 V
Described cell and standard reduction potentials
What is the standard cell potential, E°cell?
What is ΔG° for the cell reaction, Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)? (F = 96,485 C/mol e⁻)
0 of 5 answered